It is currently 21 Oct 2017, 03:35

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

What is the area of quadrilateral ABCD?

Author Message
TAGS:

Hide Tags

Intern
Joined: 27 Sep 2010
Posts: 27

Kudos [?]: 87 [0], given: 3

Show Tags

22 Jan 2011, 02:41
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

67% (00:00) correct 33% (00:06) wrong based on 16 sessions

HideShow timer Statistics

Q) In quadrilateral ABCD shown in the figure , BC || AD, AB = 8 and BC = 16. What is the area of quadrilateral ABCD?

Attachments

File comment: Figure

untitled.JPG [ 6.33 KiB | Viewed 11898 times ]

Kudos [?]: 87 [0], given: 3

Intern
Joined: 13 Oct 2010
Posts: 21

Kudos [?]: 8 [1], given: 0

Location: Milpitas, CA

Show Tags

23 Jan 2011, 09:06
1
KUDOS
Let me first redraw the figure so that it becomes more of a "to scale" figure and we can at least visualize how the actual quadrilateral will look like.

∆ABM is an isosceles right-angled triangle with angle M = 90°
Hence, BM = AM = AB/(√2) = 8/(√2) = 4√2

Hence, CN = AM = 4√2
And, AN = MC = BC - BM = (16 - 4√2)

∆CDN is a 30-60-90 triangle angle N = 90°
Hence, DN = (√3)CN = (√3)*(4√2) = 4√6

Now, area of the quadrilateral ABCD
= (area of ∆ABM) + (area of rectangle AMCN) + (area of ∆CDN)
$$= (\frac{1}{2}*AM*BM) + (AM*MC) + (\frac{1}{2}*CN*DN)$$

$$= (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{2})) + ((4\sqrt{2})*(16 - 4\sqrt{2})) + (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{6}))$$

$$= 16 + 64\sqrt{2} - 32 + 16\sqrt{3}$$

$$= 64\sqrt{2} - 16 + 16\sqrt{3}$$
Attachments

qua.jpg [ 11.53 KiB | Viewed 11736 times ]

_________________

Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)
+91-99201 32411 (India)

Last edited by Anurag@Gurome on 24 Jan 2011, 13:10, edited 1 time in total.

Kudos [?]: 8 [1], given: 0

Intern
Joined: 22 Jan 2011
Posts: 2

Kudos [?]: [0], given: 3

Show Tags

24 Jan 2011, 13:06
The image is not available...can you please post the image again?

Thanks

Kudos [?]: [0], given: 3

Intern
Joined: 21 Jan 2012
Posts: 18

Kudos [?]: 9 [0], given: 5

Show Tags

25 Sep 2012, 08:13
Anurag@Gurome wrote:
Let me first redraw the figure so that it becomes more of a "to scale" figure and we can at least visualize how the actual quadrilateral will look like.

∆ABM is an isosceles right-angled triangle with angle M = 90°
Hence, BM = AM = AB/(√2) = 8/(√2) = 4√2

Hence, CN = AM = 4√2
And, AN = MC = BC - BM = (16 - 4√2)

∆CDN is a 30-60-90 triangle angle N = 90°
Hence, DN = (√3)CN = (√3)*(4√2) = 4√6

Now, area of the quadrilateral ABCD
= (area of ∆ABM) + (area of rectangle AMCN) + (area of ∆CDN)
$$= (\frac{1}{2}*AM*BM) + (AM*MC) + (\frac{1}{2}*CN*DN)$$

$$= (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{2})) + ((4\sqrt{2})*(16 - 4\sqrt{2})) + (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{6}))$$

$$= 16 + 64\sqrt{2} - 32 + 16\sqrt{3}$$

$$= 64\sqrt{2} - 16 + 16\sqrt{3}$$

HI,

I am unable to understand how you remodeled the figure. According to your figure AB || CD. But in the question AD||BC.

Thanks!
_________________

If something helps, you must appreciate!
I'll regard
Kudos as appreciation.. Thanks

Kudos [?]: 9 [0], given: 5

Intern
Joined: 12 Jun 2012
Posts: 41

Kudos [?]: 39 [0], given: 28

Show Tags

09 Oct 2012, 08:27
Ah I was using the guys incorrect diagram - thought it was part of the questionI disagree with the answer
In quadrilateral ABCD shown in the figure , BC || AD, AB = 8 and BC = 16. What is the area of quadrilateral ABCD?

if BC is parallel to AD then the shape is a parallelogram, therefore base x height - area. (using the M/N image)

We just need height AM, which comes from a 30.60.90 triangle. The sides are therefore x.sqrt(3)x.2x
as the hypotenuse is 16, 2x = 16 so the height of the parallelogram = 8sqrt(3)
Area is therefore 8x8sqrt(3) or 64sqrt(3)

Is this right?
_________________

Kudos [?]: 39 [0], given: 28

Current Student
Joined: 06 Sep 2013
Posts: 1978

Kudos [?]: 719 [1], given: 355

Concentration: Finance

Show Tags

01 Feb 2014, 09:07
1
KUDOS
MichelleSavina wrote:
Q) In quadrilateral ABCD shown in the figure , BC || AD, AB = 8 and BC = 16. What is the area of quadrilateral ABCD?

I agree with the picture posted in the reply with the solution, but the picture posted on the original question is dead wrong, if AB =8 then MB will be 8 sqrt(2) which is larger, this doesn't make any sense. Please refer to the picture below the actual one for a more precise experience

Cheers
J

Kudos [?]: 719 [1], given: 355

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16609

Kudos [?]: 273 [0], given: 0

Show Tags

16 Sep 2016, 19:39
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Re: What is the area of quadrilateral ABCD?   [#permalink] 16 Sep 2016, 19:39
Display posts from previous: Sort by